Problem: Solve for $x$ : $6x^2 - 42x - 180 = 0$
Solution: Dividing both sides by $6$ gives: $ x^2 {-7}x {-30} = 0 $ The coefficient on the $x$ term is $-7$ and the constant term is $-30$ , so we need to find two numbers that add up to $-7$ and multiply to $-30$ The two numbers $3$ and $-10$ satisfy both conditions: $ {3} + {-10} = {-7} $ $ {3} \times {-10} = {-30} $ $(x + {3}) (x {-10}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 3) (x -10) = 0$ $x + 3 = 0$ or $x - 10 = 0$ Thus, $x = -3$ and $x = 10$ are the solutions.